\(\int \frac {A+A \sec (c+d x)}{(a-a \sec (c+d x))^{3/2}} \, dx\) [171]

Optimal result

Integrand size = 26, antiderivative size = 116 \[ \int \frac {A+A \sec (c+d x)}{(a-a \sec (c+d x))^{3/2}} \, dx=\frac {2 A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{a^{3/2} d}-\frac {3 A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{\sqrt {2} a^{3/2} d}-\frac {A \tan (c+d x)}{d (a-a \sec (c+d x))^{3/2}} \] Output:

2*A*arctan(a^(1/2)*tan(d*x+c)/(a-a*sec(d*x+c))^(1/2))/a^(3/2)/d-3/2*A*arct 
an(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a-a*sec(d*x+c))^(1/2))*2^(1/2)/a^(3/2)/ 
d-A*tan(d*x+c)/d/(a-a*sec(d*x+c))^(3/2)
 

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.09 \[ \int \frac {A+A \sec (c+d x)}{(a-a \sec (c+d x))^{3/2}} \, dx=\frac {A \left (4 \text {arctanh}\left (\sqrt {1+\sec (c+d x)}\right ) (-1+\sec (c+d x))-3 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1+\sec (c+d x)}}{\sqrt {2}}\right ) (-1+\sec (c+d x))+2 \sqrt {1+\sec (c+d x)}\right ) \tan (c+d x)}{2 a d (-1+\sec (c+d x)) \sqrt {1+\sec (c+d x)} \sqrt {a-a \sec (c+d x)}} \] Input:

Integrate[(A + A*Sec[c + d*x])/(a - a*Sec[c + d*x])^(3/2),x]
 

Output:

(A*(4*ArcTanh[Sqrt[1 + Sec[c + d*x]]]*(-1 + Sec[c + d*x]) - 3*Sqrt[2]*ArcT 
anh[Sqrt[1 + Sec[c + d*x]]/Sqrt[2]]*(-1 + Sec[c + d*x]) + 2*Sqrt[1 + Sec[c 
 + d*x]])*Tan[c + d*x])/(2*a*d*(-1 + Sec[c + d*x])*Sqrt[1 + Sec[c + d*x]]* 
Sqrt[a - a*Sec[c + d*x]])
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.29, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {3042, 4392, 3042, 4375, 373, 397, 216}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {A \sec (c+d x)+A}{(a-a \sec (c+d x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A \csc \left (c+d x+\frac {\pi }{2}\right )+A}{\left (a-a \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx\)

\(\Big \downarrow \) 4392

\(\displaystyle -a A \int \frac {\tan ^2(c+d x)}{(a-a \sec (c+d x))^{5/2}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle -a A \int \frac {\cot \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a-a \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\)

\(\Big \downarrow \) 4375

\(\displaystyle \frac {2 A \int \frac {\tan ^2(c+d x)}{(a-a \sec (c+d x)) \left (\frac {a \tan ^2(c+d x)}{a-a \sec (c+d x)}+1\right ) \left (\frac {a \tan ^2(c+d x)}{a-a \sec (c+d x)}+2\right )^2}d\left (-\frac {\tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{d}\)

\(\Big \downarrow \) 373

\(\displaystyle \frac {2 A \left (-\frac {\int \frac {1-\frac {a \tan ^2(c+d x)}{a-a \sec (c+d x)}}{\left (\frac {a \tan ^2(c+d x)}{a-a \sec (c+d x)}+1\right ) \left (\frac {a \tan ^2(c+d x)}{a-a \sec (c+d x)}+2\right )}d\left (-\frac {\tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{2 a}-\frac {\tan (c+d x)}{2 a \sqrt {a-a \sec (c+d x)} \left (\frac {a \tan ^2(c+d x)}{a-a \sec (c+d x)}+2\right )}\right )}{d}\)

\(\Big \downarrow \) 397

\(\displaystyle \frac {2 A \left (-\frac {2 \int \frac {1}{\frac {a \tan ^2(c+d x)}{a-a \sec (c+d x)}+1}d\left (-\frac {\tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )-3 \int \frac {1}{\frac {a \tan ^2(c+d x)}{a-a \sec (c+d x)}+2}d\left (-\frac {\tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{2 a}-\frac {\tan (c+d x)}{2 a \sqrt {a-a \sec (c+d x)} \left (\frac {a \tan ^2(c+d x)}{a-a \sec (c+d x)}+2\right )}\right )}{d}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {2 A \left (-\frac {\frac {3 \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{\sqrt {2} \sqrt {a}}-\frac {2 \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{\sqrt {a}}}{2 a}-\frac {\tan (c+d x)}{2 a \sqrt {a-a \sec (c+d x)} \left (\frac {a \tan ^2(c+d x)}{a-a \sec (c+d x)}+2\right )}\right )}{d}\)

Input:

Int[(A + A*Sec[c + d*x])/(a - a*Sec[c + d*x])^(3/2),x]
 

Output:

(2*A*(-1/2*((-2*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a - a*Sec[c + d*x]]])/S 
qrt[a] + (3*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a - a*Sec[c + d*x] 
])])/(Sqrt[2]*Sqrt[a]))/a - Tan[c + d*x]/(2*a*Sqrt[a - a*Sec[c + d*x]]*(2 
+ (a*Tan[c + d*x]^2)/(a - a*Sec[c + d*x])))))/d
 

Defintions of rubi rules used

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 373

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[e*(e*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 
1)/(2*(b*c - a*d)*(p + 1))), x] - Simp[e^2/(2*(b*c - a*d)*(p + 1))   Int[(e 
*x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(m - 1) + d*(m + 2*p + 
 2*q + 3)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 
 0] && LtQ[p, -1] && GtQ[m, 1] && LeQ[m, 3] && IntBinomialQ[a, b, c, d, e, 
m, 2, p, q, x]
 

rule 397

Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ 
Symbol] :> Simp[(b*e - a*f)/(b*c - a*d)   Int[1/(a + b*x^2), x], x] - Simp[ 
(d*e - c*f)/(b*c - a*d)   Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e 
, f}, x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4375

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n 
_.), x_Symbol] :> Simp[-2*(a^(m/2 + n + 1/2)/d)   Subst[Int[x^m*((2 + a*x^2 
)^(m/2 + n - 1/2)/(1 + a*x^2)), x], x, Cot[c + d*x]/Sqrt[a + b*Csc[c + d*x] 
]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[m/2] && I 
ntegerQ[n - 1/2]
 

rule 4392

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*( 
d_.) + (c_))^(n_.), x_Symbol] :> Simp[((-a)*c)^m   Int[Cot[e + f*x]^(2*m)*( 
c + d*Csc[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && E 
qQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && RationalQ[n] &&  !( 
IntegerQ[n] && GtQ[m - n, 0])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(982\) vs. \(2(99)=198\).

Time = 2.57 (sec) , antiderivative size = 983, normalized size of antiderivative = 8.47

method	result	size
default	\(\text {Expression too large to display}\)	\(983\)
parts	\(\text {Expression too large to display}\)	\(983\)

Input:

int((A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

A*(-1/8/d*2^(1/2)/a/(-a/(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^( 
1/2)*(1-cos(1/2*d*x+1/2*c))*(-8*2^(1/2)*arctanh(2^(1/2)*cos(1/2*d*x+1/2*c) 
/(cos(1/2*d*x+1/2*c)+1)/((2*cos(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c)+1) 
^2)^(1/2))*(1-cos(1/2*d*x+1/2*c))^2*csc(1/2*d*x+1/2*c)^2+5*arctanh((2*cos( 
1/2*d*x+1/2*c)-1)/(cos(1/2*d*x+1/2*c)+1)/((2*cos(1/2*d*x+1/2*c)^2-1)/(cos( 
1/2*d*x+1/2*c)+1)^2)^(1/2))*(1-cos(1/2*d*x+1/2*c))^2*csc(1/2*d*x+1/2*c)^2- 
5*ln(2*(((2*cos(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2)*cos(1/ 
2*d*x+1/2*c)+((2*cos(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2)-2 
*cos(1/2*d*x+1/2*c)-1)/(cos(1/2*d*x+1/2*c)+1))*(1-cos(1/2*d*x+1/2*c))^2*cs 
c(1/2*d*x+1/2*c)^2+((2*cos(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c)+1)^2)^( 
1/2)*(1-cos(1/2*d*x+1/2*c))^2*csc(1/2*d*x+1/2*c)^2-((2*cos(1/2*d*x+1/2*c)^ 
2-1)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2))*(cos(1/2*d*x+1/2*c)+1)^2/((2*cos(1/2 
*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2)*csc(1/2*d*x+1/2*c)^3+1/8/ 
d*(2*((2*cos(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2)*cos(1/2*d 
*x+1/2*c)+cos(1/2*d*x+1/2*c)*arctanh((2*cos(1/2*d*x+1/2*c)-1)/(cos(1/2*d*x 
+1/2*c)+1)/((2*cos(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c)+1)^2)^(1/2))-co 
s(1/2*d*x+1/2*c)*ln(2*(((2*cos(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c)+1)^ 
2)^(1/2)*cos(1/2*d*x+1/2*c)+((2*cos(1/2*d*x+1/2*c)^2-1)/(cos(1/2*d*x+1/2*c 
)+1)^2)^(1/2)-2*cos(1/2*d*x+1/2*c)-1)/(cos(1/2*d*x+1/2*c)+1))-arctanh((2*c 
os(1/2*d*x+1/2*c)-1)/(cos(1/2*d*x+1/2*c)+1)/((2*cos(1/2*d*x+1/2*c)^2-1)...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 211 vs. \(2 (99) = 198\).

Time = 0.10 (sec) , antiderivative size = 502, normalized size of antiderivative = 4.33 \[ \int \frac {A+A \sec (c+d x)}{(a-a \sec (c+d x))^{3/2}} \, dx=\left [-\frac {3 \, \sqrt {2} {\left (A \cos \left (d x + c\right ) - A\right )} \sqrt {-a} \log \left (\frac {2 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} + {\left (3 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )}{{\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) + 4 \, {\left (A \cos \left (d x + c\right ) - A\right )} \sqrt {-a} \log \left (\frac {2 \, {\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} - {\left (2 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )}{\sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 4 \, {\left (A \cos \left (d x + c\right )^{2} + A \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}}}{4 \, {\left (a^{2} d \cos \left (d x + c\right ) - a^{2} d\right )} \sin \left (d x + c\right )}, \frac {3 \, \sqrt {2} {\left (A \cos \left (d x + c\right ) - A\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 4 \, {\left (A \cos \left (d x + c\right ) - A\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) + 2 \, {\left (A \cos \left (d x + c\right )^{2} + A \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}}}{2 \, {\left (a^{2} d \cos \left (d x + c\right ) - a^{2} d\right )} \sin \left (d x + c\right )}\right ] \] Input:

integrate((A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(3/2),x, algorithm="fricas")
 

Output:

[-1/4*(3*sqrt(2)*(A*cos(d*x + c) - A)*sqrt(-a)*log((2*sqrt(2)*(cos(d*x + c 
)^2 + cos(d*x + c))*sqrt(-a)*sqrt((a*cos(d*x + c) - a)/cos(d*x + c)) + (3* 
a*cos(d*x + c) + a)*sin(d*x + c))/((cos(d*x + c) - 1)*sin(d*x + c)))*sin(d 
*x + c) + 4*(A*cos(d*x + c) - A)*sqrt(-a)*log((2*(cos(d*x + c)^2 + cos(d*x 
 + c))*sqrt(-a)*sqrt((a*cos(d*x + c) - a)/cos(d*x + c)) - (2*a*cos(d*x + c 
) + a)*sin(d*x + c))/sin(d*x + c))*sin(d*x + c) - 4*(A*cos(d*x + c)^2 + A* 
cos(d*x + c))*sqrt((a*cos(d*x + c) - a)/cos(d*x + c)))/((a^2*d*cos(d*x + c 
) - a^2*d)*sin(d*x + c)), 1/2*(3*sqrt(2)*(A*cos(d*x + c) - A)*sqrt(a)*arct 
an(sqrt(2)*sqrt((a*cos(d*x + c) - a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*s 
in(d*x + c)))*sin(d*x + c) - 4*(A*cos(d*x + c) - A)*sqrt(a)*arctan(sqrt((a 
*cos(d*x + c) - a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c)))*sin( 
d*x + c) + 2*(A*cos(d*x + c)^2 + A*cos(d*x + c))*sqrt((a*cos(d*x + c) - a) 
/cos(d*x + c)))/((a^2*d*cos(d*x + c) - a^2*d)*sin(d*x + c))]
 

Sympy [F]

\[ \int \frac {A+A \sec (c+d x)}{(a-a \sec (c+d x))^{3/2}} \, dx=A \left (\int \frac {\sec {\left (c + d x \right )}}{- a \sqrt {- a \sec {\left (c + d x \right )} + a} \sec {\left (c + d x \right )} + a \sqrt {- a \sec {\left (c + d x \right )} + a}}\, dx + \int \frac {1}{- a \sqrt {- a \sec {\left (c + d x \right )} + a} \sec {\left (c + d x \right )} + a \sqrt {- a \sec {\left (c + d x \right )} + a}}\, dx\right ) \] Input:

integrate((A+A*sec(d*x+c))/(a-a*sec(d*x+c))**(3/2),x)
 

Output:

A*(Integral(sec(c + d*x)/(-a*sqrt(-a*sec(c + d*x) + a)*sec(c + d*x) + a*sq 
rt(-a*sec(c + d*x) + a)), x) + Integral(1/(-a*sqrt(-a*sec(c + d*x) + a)*se 
c(c + d*x) + a*sqrt(-a*sec(c + d*x) + a)), x))
 

Maxima [F]

\[ \int \frac {A+A \sec (c+d x)}{(a-a \sec (c+d x))^{3/2}} \, dx=\int { \frac {A \sec \left (d x + c\right ) + A}{{\left (-a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(3/2),x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

integrate((A*sec(d*x + c) + A)/(-a*sec(d*x + c) + a)^(3/2), x)
 

Giac [A] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.97 \[ \int \frac {A+A \sec (c+d x)}{(a-a \sec (c+d x))^{3/2}} \, dx=\frac {\frac {3 \, \sqrt {2} A \arctan \left (\frac {\sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a}}{\sqrt {a}}\right )}{a^{\frac {3}{2}}} - \frac {4 \, A \arctan \left (\frac {\sqrt {2} \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a}}{2 \, \sqrt {a}}\right )}{a^{\frac {3}{2}}} - \frac {\sqrt {2} \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a} A}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}{2 \, d} \] Input:

integrate((A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(3/2),x, algorithm="giac")
 

Output:

1/2*(3*sqrt(2)*A*arctan(sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)/sqrt(a))/a^(3/2 
) - 4*A*arctan(1/2*sqrt(2)*sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)/sqrt(a))/a^( 
3/2) - sqrt(2)*sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)*A/(a^2*tan(1/2*d*x + 1/2 
*c)^2))/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+A \sec (c+d x)}{(a-a \sec (c+d x))^{3/2}} \, dx=\int \frac {A+\frac {A}{\cos \left (c+d\,x\right )}}{{\left (a-\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \] Input:

int((A + A/cos(c + d*x))/(a - a/cos(c + d*x))^(3/2),x)
 

Output:

int((A + A/cos(c + d*x))/(a - a/cos(c + d*x))^(3/2), x)
 

Reduce [F]

\[ \int \frac {A+A \sec (c+d x)}{(a-a \sec (c+d x))^{3/2}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {-\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{2}-2 \sec \left (d x +c \right )+1}d x +\int \frac {\sqrt {-\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )}{\sec \left (d x +c \right )^{2}-2 \sec \left (d x +c \right )+1}d x \right )}{a} \] Input:

int((A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(3/2),x)
 

Output:

(sqrt(a)*(int(sqrt( - sec(c + d*x) + 1)/(sec(c + d*x)**2 - 2*sec(c + d*x) 
+ 1),x) + int((sqrt( - sec(c + d*x) + 1)*sec(c + d*x))/(sec(c + d*x)**2 - 
2*sec(c + d*x) + 1),x)))/a